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I'm trying to derive the electric field due to a single large, thin, non-conducting plate at a point (see figure). I'm solving it using 2 methods, and arriving at a different answer using both.

I've referred some textbooks, and they say that the result of the 2nd derivation is correct. I would like to know which method is correct, and why is the other method wrong? Can I change any equation/assumption in the wrong method to arrive at the right result?

Derivation 1:

Derivation 2:

enter image description here

Images produced by myself using this website.

BlazeRod11
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  • Possible duplicates: https://physics.stackexchange.com/q/65191/2451 and links therein. – Qmechanic Jun 07 '21 at 19:29
  • Thanks for the reply @Qmechanic. I had read that thread before posting but was unable to find the exact reason as to why the Gauss Law application in the 1st derivation was incorrect. – BlazeRod11 Jun 07 '21 at 19:37

2 Answers2

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The first derivation is incorrect because we assume the sheet of charge to be infinitely thin and the surface you are using to apply Gauss Law is also infinitely thin, and so the Gaussian surface must either contain the charged sheet (as it does in derivation 2), or it doesn't contain the second sheet, in which case $Q_{enc}=0$ and so Gauss Law doesn't do anything for us, since we just get $0=0$.

As you mention in the question the second derivation is what gives us the correct answer for Electric Field due to this large-thin sheet, and is how its done in most all textbooks.

xXx_69_SWAG_69_xXx
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  • Thanks for the answer, @xXx_69_SWAG_69_xXx! I have another query. If the plate were a conducting plate (part of a capacitor), would it still be valid to consider the effect of the electric field due to the left side, on any point towards the right in derivation 2 (since electric field does not exist within the volume of the conductor, and therefore cannot propagate through it)? – BlazeRod11 Jun 07 '21 at 19:51
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Since it's a nonconducting plate, the charge sits only on the left surface and there is indeed an electric field inside the material (we're ignoring dielectric effects here, right? otherwise you'll need to know the dielectric constant of the material.) So I would say that your mistake is that you did NOT draw the electric field going to the right inside the material in your first figure (Derivation 1). You can keep the Gaussian surface inside the material, but there IS an electric field in there, just as you've drawn in the Derivation 2. The charge enclosed is the same in both pictures, and the flux is 2EA in both pictures.

Note also that if this were a conductor, then the electric field would be zero inside the material and Derivation 1 gives the correct answer. Of course, if it were a conductor, then there must be an equal amount of charge on the right surface of the conducting plate. This would give E = 0 inside, and $E = \sigma/\epsilon_0$ outside

quantumwave
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  • Really nice explanation! So if it were a conducting plate, can we say that each side of the plate produces an electric field E = σ/2ϵ0, and that the net E at any point will be equal Enet = σ/2ϵ0 + σ/2ϵ0 (since both sides produce an outward electric field?) – BlazeRod11 Jun 09 '21 at 10:46
  • That's exactly right for regions outside the conducting plate. Inside the plate, the field contributions cancel $\vec{E}_{in} = \frac{\sigma}{2\epsilon_0}\hat{x} - \frac{\sigma}{2\epsilon_0}\hat{x} = 0$. – quantumwave Jun 10 '21 at 14:35